electrical machines ii lab manual · the scott-t connection is the method of connecting two single...

ELECTRICAL MACHINES – II

LAB MANUAL

ELECTRICAL AND ELECTRONICS ENGINEERING

ANURAG COLLEGE OF ENGINEERING

Aushapur (V), Ghatkesar (M) - 501301

ELECTRICAL MACHINES - II LABORATORY

ANURAG COLLEGE OF ENGINEERING, ANRH DEPARTMENT OF EEE

DEPARTMENT OF ELECTRICAL AND ELECTRONICS

ENGINEERING

ELECTRICAL MACHINES-II LABARATORY

LIST OF EXPERIMENTS

Sl.No Name of the Experiment

1 OPEN CIRCUIT & SHORT CIRCUIT TEST ON A SINGLE PHASE

TRANSFORMER

2 SUMPNER'S TEST ON SINGLE PHASE TRANSFORMERS

3 SCOTT CONNECTION OF TWO 1-PH TRANSFORMERS

4 NO LOAD AND BLOCKED ROTOR TEST ON A 3- ɸ

INDUCTION MOTOR

5 REGULATION OF ALTERNATOR USING SYNCHRONOUS

IMPEDANCE METHOD

6 ‘V’ AND ‘INVERTED V’ CURVES OF SYNCHRONOUS MOTOR

7 EQUIVALENT CIRCUIT OF A SIGLE PHASE INDUCTION

MOTOR

8 BRAKE TEST ON 3- ɸ SQUIRREL CAGE INDUCTION MOTOR

9 SEPARATION OF NO LOAD LOSSES IN 1-Φ TRANSFORMER

10 DETERMINATION OF Xd AND Xq OF SALIENT POLE

SYNCHRONOUS MOTOR

ANURAG COLLEGE OF ENGINEERINGELECTRICAL MACHINES - II LABORATORY


OPEN CIRCUIT & SHORT CIRCUIT TESTA ON

A SINGLE PHASE TRANSFORMER

AIM:

To perform open circuit and short circuit test on a single phase transformer and draw

its equivalent circuit, to determine the efficiency at different loads and power factors, to

find the load at whichi maximum efficiency occurs, to calculate all day efficiency.

APPARATUS REQUIRED:

Sl.

No. equipment Type Range Quantity

1 Voltmeter MI

(0-300)V

(0-150)V

1 no

1 no

2 Ammeter MI

(0-1/2)A

(0-10/20)A

1 no

1 no

3 Wattmeter Dynamo type

(0-150)V LPF

(0-1)A

1 no


(0-150)V UPF

(0-10)A

1 no

5 Connecting Wires ***** ***** Required

NAME PLATE DETAILS:

Rating:

Primary Voltage:

Secondary Voltage:



CIRCUIT DIAGRAM:

OPEN CIRCUIT:

SHORT CIRCUIT:

THEORY:

The efficiency and regulation of a transformer on any load condition and at any power factor condition can be predetermined by indirect loading method. In this method, the actual load is not used on transformer. But the equivalent circuit parameters of a transformer are determined by conducting two tests on a transformer which are,1. Open circuit test2. Short circuit test

The parameters calculated from these test results are effective in determining the regulation and efficiency of a transformer at any load and power factor condition, without actually loading the transformer. The advantage of this method is that without much power loss the tests can be performed and results can be obtained.



PROCEDURE:

Open circuit test:

1. Connections are made as per the circuit diagram.

2. Ensure that variac is set to zero output voltage position before starting the

experiment.

3. Switch ON the supply. Now apply the rated voltage to the Primary winding by

using Variac.

4. The readings of the Voltmeter, ammeter and wattmeter are noted down in Tabular

form.

5. Then Variac is set to zero output position and switch OFF the supply.

6. Calculate Ro and Xo from the readings.

Short Circuit Test:


2. Ensure that variac is set to zero output voltage position before starting the

experiment.

3. Switch ON the supply. Now apply the rated Current to the Primary winding by

using Variac.

4. The readings of the Voltmeter, ammeter and wattmeter are noted down in Tabular

form.

5. Then Variac is set to zero output position and switch OFF the supply.

6. Calculate Ro1 and Xo1 from the readings.

OBSERVATIONS:

I) For OC test

Sl no.

Voltmeter

reading

( Vo)

Ammeter

reading

(Io)

Wattmeter

reading

Wo

Ro Xo Cos ɸo



II) For SC test

Sl no.

Voltmeter

reading

( VSC)

Ammeter

reading

(ISC)

Wattmeter

reading

WSC

Ro1 Zo1 Xo1

MODEL CALCULATIONS:

Find the equivalent circuit parameters R0, X0, R01, R02, X01 and X02 from the O. C. and S.

C. test results and draw the equivalent circuit referred to L. V. side as well as H. V. side.

Let the transformer be the step-down transformer

Primary is H. V. side.

Secondary is L. V. side wI

VR 1

0 where Iw = I0 cos 0

mI

VX 1

0 Where Im = I0 sin 0

SC

SC

sc

SC

I

VZ

I

WR 0120 ,

1

2

01

2

0101RZX : 01

2

02 XKX Where K = 1

2

V

V Transformation ratio.

Calculations to find efficiency and regulation

For example at ½ full load

Cupper losses = Wsc x (1/2)2 watts, where WSC = full – load cupper losses

Constant losses = W0 watts

Output = ½ KVA x cos [cos may be assumed]

Input = output + Cu. Loss + constant loss

% 100xInput

Outputefficiency

Efficiency at different loads and P.f’s



cos = ___________

Regulation: From open circuit and Short circuit test

100sincos

Re%2

022022 xV

XIRIgulation

‘+’ for lagging power factors

‘-‘ for leading power factor

S.No p.f. % reg

Lag Lead

Cos = 1.0

S.No Load Wcu (W) O/P (W) I/P (W) (%)

Cos = 0.8

S.No Load Wcu (W) O/P (W) I/P (W) (%)



GRAPHS: Plots drawn between

(i) % efficiency Vs output

(ii) % Regulation Vs Power factor

PRECAUTIONS:

(i) Connections must be made tight

(ii) Before making or breaking the circuit, supply must be switched off

RESULT:

CONCLUSION:

VIVA QUESTIONS:

1. Which losses are called magnetic losses?

2. Write equations for hysteresis and eddy - current losses.

3. Why O.C test is conducted on LV side?

4.

5. Why transformer fails to operate on D.C supply?

6. Explain why low power factor meter is used in O.C test.

Why S.C test will conducted on HV side?



SUMPNER'S TEST ON A PAIR OF TWO SINGLE PHASE TRANSFORMERS

AIM:

To determine the efficiency and losses of a given transformer accurately under

full load condition.

APPARATUS REQUIRED:

Sl.

No. Equipment Type Range Quantity

1 Voltmeter MI

(0-300)V

(0-300)V

(0-600)V

1 no

1 no

1 no

2 Ammeter MI

(0-2)A

(0-20)A

1 no

1 no


(0-150)V LPF

(0-2)A

1 no


(0-150)V UPF

(0-10)A

1 no


NAME PLATE DETAILS:

Rating

Primary Voltage

Secondary Voltage



CIRCUIT DIAGRAM:

PROCEDURE:

1. Make the connections as per the circuit diagram.

2. The secondary winding terminals of the two transformers are connected in series

with polarities in phase opposition which can checked by means of a voltmeter.

THEORY:

Sumpner's test or back to back test on transformer is another method for determining transformer efficiency, voltage regulation and heating under loaded conditions. Short circuit and open circuit tests on transformer can give us parameters of equivalent circuit of transformer, but they can not help us in finding the heating information. Unlike O.C. and S.C. tests, actual loading is simulated in Sumpner's test. Thus the Sumpner's test give more accurate results of regulation and efficiency than O.C. and S.C. tests. Sumpner's test or back to back test can be employed only when two identical transformers are available. Both transformers are connected to supply such that one transformer is loaded on another. Primaries of the two identical transformers are connected in parallel across a supply. Secondaries are connected in series such that emf's of them are opposite to each other. Another low voltage supply is connected in series with secondaries to get the readings



3. Before starting the experiment, check the variacs are in minimum output voltage

position.

4. Close the first DPST-1 switch and switch ON the supply.

5. Increase the variac slowly, and apply rated voltage to the primary windings of 1-

ɸ transformers and check the voltmeter reading connected across the secondary

terminals.

6. If the voltmeter reading is Zero, continue with step 8.

7. If the voltmeter reading is not zero, interchange the secondary terminals.

8. Now close the DPST-2 switch and vary the variac-2 slowly till rated current flows

in the two series-connected secondaries.

9. Note down the readings of V1,V2, I1, I2, W1, and W2 and enter them in a tabular

column.

10. W1 = 2Pc, W2= 2Psc. Losses of each transformer = (W1+W2)/2

11. Now the Variacs are brought to zero voltage position and open DPST switches.

OBSERVATIONS:

Sl

no.

Voltmeter

reading

V1

Voltmeter

reading

V2

Ammeter

reading

I1

Ammeter

reading

I1

Wattmeter

Reading

W1

Wattmeter

Reading

W2

Transformer

losses

= (W1+W2)/2

ɳ =

output

(op+loss)

MODEL CALCULATIONS:

Losses in each transformer = 2

ci ww % combined = 100

1

1 xwwIV

IV

ci



Efficiency of each transformer (% )= 100

221

1 xww

IV

IV

ci

MODEL GRAPH:

Output power Vs Efficiency

PRECAUTIONS:

1. Connections must be made tight

2. Before making or breaking the circuit, supply must be switched off

RESULT:

CONCLUSION:

VIVA QUESTIONS:

1. What for this test is really intended?

2. Why to conduct the test on identical transformers?

3. What happens if the rated values of voltage and frequency of supply vary?

4. What are the advantages and disadvantages of this test?

5. Can you perform this test on 3 – star/ delta transformers?

6. What is all-day efficiency?



SCOTT CONNECTION OF TRANSFORMERS

AIM:

To perform the Scott connection of transformer and to obtain the two phase

supply from three phase supply

APPARATUS REQUIRED:

Sl.


1 Voltmeter MI

(0-300)V

(0-600)V

2 no

2 no

2 Ammeter MI (0-10)A 3 no


NAME PLATE DETAILS:

Rating

Primary Voltage

Secondary Voltage

THEORY:

The Scott-T Connection is the method of connecting two single phase transformer to perform the 3-phase to 2-phase conversion and vice-versa. The two transformers are connected electrically but not magnetically.One of the transformers is called the main transformer, and the other is called the auxiliary or teaser transformer.



CIRCUIT DIAGRAM:

PROCEDURE:

1. Connections are made as per the circuit diagram

2. Ensure that output voltage of the variac is set in zero position before starting the

experiment.

3. Switch ON the supply.



4. The output voltage of the variac is gradually increased in steps upto rated voltage

of single phase MAIN transformer and readings are correspondingly taken in

steps.

5. Enter the readings in tabular column.

6. After observations, the variac is brought to zero position and switch OFF the

supply.

CALCULATIONS:

Prove

TABULAR COLUMN:

Sl

no.

Voltmeter

reading

V1

Ammeter

reading

I1

Voltmeter

reading

V2T

Voltmeter

reading

V2M

Voltmeter

reading

V2TM

Theoretical

calculation

V2TM =√ (V2

2T +

V2

2M)

RESULT:

CONCLUSION:

1. What is the use of Scott connection?

2. Compare open delta, Scott connections.

3. How the Scott connection is formed?

4. One transformer has cruciform type and second transformer has square type of core which is the

better one?

5. Draw the phasor diagram for Scott connection.

VIVA QUESTIONS:



NO LOAD AND BLOCKED ROTOR TEST ON A 3- ɸ

INDUCTION MOTOR

AIM:

To determine the equivalent circuit of a 3- ɸ induction motor and calculate

various parameters of induction motor with the help of circle diagram.

APPARATUS REQUIRED:

Sl.


1 Voltmeter MI (0-600)V 1 no


3 Wattmeter Electro dynamo meter type

10A/600V UPF

10A/600V LPF

1 no

1 no

4 Tachometer Digital 1-10000 RPM 1 no


NAME PLATE DETAILS:

Power rating

Voltage

Current

Speed(RPM)



CIRCUIT DIAGRAM:

PROCEDURE:

NO LOAD TEST:


THEORY:

An electrical motor is such an electromechanical device which converts electrical energy into a mechanical energy. In case of three phase AC operation, most widely used motor is Three phase induction motor as this type of motor does not require any starting device or we can say they are self-starting induction motors.

When a three phase supply is give to the stator, due to interaction of flux produced by the three phase current flowing through the stator , RMF at synchronous speed is produced, as a result EMF is induced in the rotor bars (for squirrel cage) and as a result of this current will be flowing in the rotor conductors. Conductor placed in a magnetic field will experience a force. As per Lenz's law the rotor will rotate in the direction of Rotating Magnetic Field to oppose the effect. This causes the motor to rotate.



2. Ensure that the 3- ɸ variac is kept at minimum output voltage position and belt is

freely suspended.

3. Switch ON the supply. Increase the variac output voltage gradually until rated

voltage is observed in voltmeter. Note that the induction motor takes large current

initially, so, keep an eye on the ammeter such that the starting current current

should not exceed 7 Amp.

4. By the time speed gains rated value, note down the readings of voltmeter,

ammeter, and wattmeter.

5. Bring back the variac to zero output voltage position and switch OFF the supply.

BLOCKED ROTOR TEST:

1. Connections are as per the circuit diagram.

2. The rotor is blocked by tightening the belt.

3. A small voltage is applied using 3- ɸ variac to the stator so that a rated current

flows in the induction motor.

4. Note down the readings of Voltmeter, Ammeter and Wattmeter in a tabular

column.

5. Bring back the Variac to zero output voltage position and switch OFF the supply.

OBSERVATIONS:

No Load Test:

Sl no. Voltmeter

reading

V nl

Ammeter

reading

Inl

Wattmeter reading Wnl (Pnl)

W1+W2 W1 W2



Blocked Rotor Test

Sl no. Voltmeter

reading

V br

Ammeter

reading

Ibr

Wattmeter reading Wbr (Pbr)

W1+W2 W1 W2

Measurement of stator winding resistance (r1):

CIRCUIT DIAGRAM:

TABULAR COLUMN:

S no. Voltage (v) Ammeter (I) Resistance (R)

Procedure to find r 1:

1. Connections are made as per the circuit diagram

2. Switch ON the supply. By varying the rheostat, take different readings of

ammeter and voltmeter in a tabular column.



3. From the above readings, average resistance r1 of a stator is found

Measurement of Stator resistance

1. Connect the circuit as per the circuit diagram shown in fig (2).

2. Keeping rheostat in maximum resistance position switch on the 220 V Dc supply.

3. Using volt-ammeter method measure the resistance of the stator winding.

4. After finding the stator resistance, Rdc must be multiplied with 1.6 so as to

account for skin effect i.e. Rac = 1.6 Rdc.

MODEL CALCULATIONS:

2

0

2

00

0

02

0

0 ,,3

GYBV

IY

V

WG

2

01

2

010120110 ,3

, RZXIx

WR

I

VZ

SC

SC

SC

SC

For circle diagram

00

01

0

00

0

03

cos,3

cosIV

W

IV

W

CS

CSNS

CSSC

CS

V

VII

IV

W0

0 ,3

cos

PRECAUTIONS:



RESULT:

CONCLUSION:



VIVA QUESTIONS:

1. Explain why the locus of the induction motor current is a circle.

2. What is the difference between the transformer equivalent circuit and induction

motor equivalent circuit?

3. What are the reasons in conducting no-load test with rated voltage and blocked-

rotor test with rated current?

4. Why do you choose LPF wattmeter in load test and hpf wattmeter in blocked rotor

test?

5. How do you reverse the direction of rotation of induction motor?

6. What are the various applications of this motor?



REGULATION OF ALTERNATOR USING SYNCHRONOUS IMPEDANCE METHOD

AIM:

To find the regulation of a 3 - ɸ alternator by using synchronous impedance

method.

APPARATUS REQUIRED:

Sl.


1 Voltmeter MI (0-300/600)V 1 no

2 Ammeter MI (0-5/10)A 1 no

3 Ammeter MI (0-1/)A 1 no

3 Rheostat Wire-wound

350 Ω /1.7A

350 Ω /1.7A

1 no

1 no



NAME PLATE DETAILS:

DC Motor(prime mover) 3- ɸ Alternator

Rating : Power Rating:

Voltage : PF :

Current : Line voltage:

Speed : Speed

Exctn : Shunt Exctn Voltage:

Rated Current :



CIRCUIT DIAGRAM:

PROCEDURE:

Open Circuit Test:

1. Make the connections as per the circuit diagram.

2. Before starting the experiment, the potential divider network in the alternator field

circuit and field regulator rheostat of motor circuit is set minimum resistance

position.

3. Switch ON the supply and close the DPST switch. The DC motor is started by

moving starter handle.

4. Adjust the field rheostat of DC motor to attain rated speed (equal to synchronous

speed of alternator)

5. By decreasing the field resistance of Alternator, the excitation current of

alternator is increased gradually in steps.

THEORY:

An alternator is an electrical generator that converts mechanical energy to electrical energy in the form of alternating current. For reasons of cost and simplicity, most alternators use a rotating magnetic field with a stationary armature. Occasionally, a linear alternator or a rotating armature with a stationary magnetic field is used. In principle, any AC electrical generator can be called an alternator, but usually the term refers to small rotating machines driven by automotive and other internal combustion engines.



6. Note the readings of field current, and its corresponding armature voltage in a

tabular column.

7. The voltage readings are taken upto and 10% beyond the rated voltage of the

machine.

Short Circuit Test:

1. For Short circuit test, before starting the experiment the potential divider is

brought back to zero output position, i.e., resistance should be zero in value.

2. Now close the TPST switch.

3. The excitation of alternator is gradually increased in steps until rated current

flows in the machine and note down the readings of excitation current and load

current (short circuit current)

4. Switch OFF the supply.

OBSERVATIONS:

Sl

no.

OC test Sl

no.

S.C. test

Field current in

Amp.(I f)

OC voltage

per phase (Vo)

Field current

If( Amp.)

SC current

Isc Amp.

Procedure to find Armature resistance of alternator:


2. Switch ON the supply. By varying the rheostat, take different readings of

ammeter and voltmeter in a tabular column.



3. From the above readings, average resistance Ra of a armature is found out.

Connection diagram to find Ra

OBSERVATIONS:

Sl no. Armature current

I(amp)

Armature voltage

Va (volts)

Rdc=V / I

Procedure to find synchronous impedance from OC and SC tests:

1. Plot open circuit voltage, short circuit current verses field current on a graph

sheet.

2. From the graph, the synchronous impedance for the rated value of excitation is

calculated.

3. The excitation emf is calculated at full load current which is equal to the terminal

voltage at No load.



4. The voltage regulation is calculated at rated terminal voltage.

MODEL CALCULATIONS:

SC

OC

SI

VZ for the same If and speed: 22

aSS RZX [Ra RdC]

Generated emf of alternator on no load is

22

0 sincos Saaa XIvRIvE

+ for lagging p.f.

- for leading p.f.

The percentage regulation of alternator for a given p.f. is

100Re% 0 xV

VEg

Where

E0 – generated emf of alternator (or excitation voltage per phase)

V – full load, rated terminal voltage per phase.

MODEL GRAPHS:

Draw the graph between If VS E0 per phase



and If VS ISC

PRECAUTIONS:

(iii) Connections must be made tight

(iv) Before making or breaking the circuit, supply must be switched off

RESULT:

CONCLUSION:

1. What is regulation of alternator?

2. Under what condition, regulation is positive or negative?

3. What is regulation at UPF

4. Why Regulation is so important in alternator?

5. How the regulation is effected by armature reaction?

VIVA QUESTIONS:



‘V’ AND ‘INVERTED V’ CURVES OF SYNCHRONOUS

MOTOR

AIM:

To plot the ‘v’ and ‘inverted v’ curves of Synchronous motor.

APPARATUS REQUIRED:

Sl.



2 Ammeter

MC

MI

(0-1/2)A

(0-10/20)A

1 no

1 no

3 Rheostat Wire-wound 350 Ω /1.7A 1 no


5 Wattmeter Electrodynamometer

10A, 600V UPF

10A , 600V LPF

1 no

1 no


NAME PLATE DETAILS

3- ɸ Synchronous motor

Power Rating:

Rated voltage

rated current:

Speed



CIRCUIT DIAGRAM:

PROCEDURE:


THEORY:

Definition: The motor which runs at synchronous speed is known as the synchronous motor. The synchronous speed is the constant speed at which motor generates the electromotive force. The synchronous motor is used for converting the electrical energy into mechanical energy.

The stator and the rotor are the two main parts of the synchronous motor. The stator becomes stationary, and it carries the armature winding of the motor. The armature winding is the main winding because of which the EMF induces in the motor. The rotator carry the field windings. The main field flux induces in the rotor. The rotor is designed in two ways, i.e., the salient pole rotor and the non-salient pole rotor.

T he synchronous motor uses the salient pole rotor. The word salient means the poles of the rotor projected towards the armature windings. The rotor of the synchronous motor is made with the laminations of the steel. The laminations reduce the eddy current loss occurs on the winding of the transformer. The salient pole rotor is mostly used for designing the medium and low-speed motor. For obtaining the high-speed cylindrical rotor is used in the motor.



2. Opening the SPST switch connected across the field DC supply is given to the

field and field current is adjusted to 0.3A ( 20% of rated field current)

3. The DC supply to the field is removed and SPST switch is connected across the

field by closing the switch

4. As 3- ɸ , 440V, 50Hz AC supply is applied to 3- ɸ dimmer stator keeping it in

minimum output position, keeping it prior to that motor is kept in no load state.

5. Gradually supply voltage to synchronous motor is increased and then motor starts

running as squirrel cage induction motor. The direction of rotation is observed. if

it is not proper then supply phase sequence is altered.

6. Observing Ia, the voltage is gradually increased. It will reach a high value and

suddenly falls to a low value.

7. At that instant, open SPST switch connected across the field. The DC supply is

then given to the field. Then the motor is pulled into synchronism and motor now

works as a synchronous motor.

8. Gradually the supply voltage to stator is increased by observing the armature

current. If Ia, increases above the rated value then increase If such that Ia will be

within limits and thus full rated supply voltage is gradually given to the motor.

Now motor will work as synchronous motor with full rated voltage.

9. By varying If in steps, armature currents are recorded at no-load.

10. By applying half of full load on motor, If and Ia are recorded again. The same

experiment is repeated at 3/4th load, full load and corresponding readings are

recorded.

11. Completely removing the load on motor, the 3- ɸ supply to stator and then the

DC supply to the field are switched OFF

OBSERVATION TABLE:

Sl

no.

Supply

voltage

Wattmeter

W1

Wattmeter

W2

Field current

If(Amp)

Armature current

Ia(Amp)

Cos ɸ



31

Load 1 : 18.1% FL N = 1500 rpm VL = 415V S1 = 2.2 kg S2 = 5.2 lg

If

(A)

Ia

(A)

W1

(W)

W2

(W)

21

213

WW

WWTan

Cos

Load 2 : 39.2% FL N = 1500 rpm VL = 415V S1 = 3.5 kg S2 = 9 kg

If

(A)

Ia

(A)

W1

(W)

W2

(W)

21

213

WW

WWTan

Cos

CALCULATIONS:

Power factor =Cos [tan-1 ]

21

211 3

WW

WWTan

MODEL GRAPHS:

21

213

WW

WW



RESULT:

VIVA Questions:

1. What are the difficulties in starting a synchronous motor?

2. What are the commonly employed methods of starting a synchronous motor?

3. What are the applications of synchronous motor?

4. What is synchronous condenser?

5. What do you understand by hunting?

CONCLUSION:



EQUIVALENT CIRCUIT OF A SIGLE PHASE INDUCTION

MOTOR

AIM:

To determine the equivalent circuit parameters of a single phase induction motor

by performing the

no- load and blocked rotor tests.

APPARATUS REQUIRED:

Sl.




3 Wattmeter Dynamo-type

(0-300)V LPF

(0-10)A

1 no

4 Wattmeter Dynamo-type

(0-150)V UPF

(0-10)A

1 no


NAME PLATE DETAILS

1- ɸ Induction motor

Power Rating:

Rated voltage

rated current:

Speed



CIRCUIT DIAGRAM:

PROCEDURE:

No load Test:

1. The circuit connections are made as per the circuit diagram.

2. Be sure that variac (auto transformer) is set to zero output voltage position before

starting the experiment.

3. Now switch ON the supply and close the DPST switch.

4. The variac is varied slowly, until rated voltage is applied to motor and rated speed

is obtained.

5. Take the readings of Ammeter, Voltmeter and wattmeter in a tabular column.

THEORY:

A Single Phase Induction Motor consists of a single phase winding which is mounted on the stator of the motor and a cage winding placed on the rotor. A pulsating magnetic field is produced, when the stator winding of the single-phase induction motor shown below is energised by a single phase supply.

The word Pulsating means that the field builds up in one direction falls to zero and then builds up in the opposite direction. Under these conditions, the rotor of an induction motor does not rotate. Hence, a single phase induction motor is not self-starting. It requires some special starting means.



6. The variac is brought to zero output voltage position after the experiment is done,

and switch OFF the supply.

Blocked Rotor Test:

1. To conduct blocked rotor test, necessary meters are connected to suit the full load

conditions of the motor.


3. Before starting the experiment variac (auto transformer) is set to zero output

voltage position.

4. The rotor (shaft) of the motor is held tight with the rope around the brake drum.

5. Switch ON the supply, and variac is gradually varied till the rated current flows in

the induction motor.

6. Readings of Voltmeter, Ammeter, and wattmeter are noted in a tabular column.

7. The variac is brought to zero output voltage position after the experiment is done,

and switch OFF the supply.

8. Loosen the rope after the experiment is done.

Calculation for No-Load Test:



Calculation For Blocked Rotor Test:

OBSERVATIONS:

For NO-Load Test:

Sl no. Voltmeter reading

Vo

Ammeter reading

Io

Wattmeter reading

Wo

For Blocked Rotor Test:

Sl no. Voltmeter reading

Vsc

Ammeter reading

Isc

Wattmeter reading

Wsc

PROCEDURE:




2. Initially rheostat is set at maximum resistance position.

3. Switch ON the supply, and vary the rheostat gradually and note down the readings

of ammeter and voltmeter

4. For the corresponding values, average of r1 is taken.

Circuit diagram for measurement of R1 :

To find stator Resistance:

S.No I (A) V (volts) R =

I

V ()

Average Value: Rdc Rac Rdc =



Comments:

1. Since IM is not self starting Machine, it is started by placing an auxiliary winding

in the circuit.

2. Here no-load test is similar to open circuiting the load terminals and blocking the

rotor is similar to conducting short circuit on the IM.

PRECAUTIONS:

Connections must be made tight

Before making or breaking the circuit, supply must be switched off

CONCLUSION:

RESULT:

VIVA QUESTIONS:

1. Why there is no starting torque in a single phase induction motor?

2. What are different starting methods employed in single phase induction motors?

3. Compare the performance of capacitor - start, capacitor – run, shaded pole single

phase induction motors?

4. Mention a few applications of single phase induction motors?



BRAKE TEST ON 3- ɸ SQUIRREL CAGE INDUCTION

MOTOR

AIM:

To determine the efficiency of 3- ɸ induction motor by performing load test. To

obtain the performance curves for the same.

APPARATUS REQUIRED:

Sl.




3 Wattmeter Electro dynamo meter type

10A/0-600V UPF

10A/0-600V LPF

1 no

1 no



NAME PLATE DETAILS

3- ɸ squrrel cage induction motor

Power Rating:

Rated voltage

rated current:

Speed



CIRCUIT DIAGRAM:

PROCEDURE:


2. Ensure that the 3- ɸ variac is kept at minimum output voltage position and belt is

freely suspended.

THEORY:

A squirrel-cage rotor is the rotating part (rotor) used in the most common form of AC induction motor. It consists of a cylinder of steel with aluminum or copper conductors embedded in its surface. An electric motor with a squirrel-cage rotor is termed a squirrel-cage motor. Basically when the 3 phase supply is applied to the 3 windings, 1 phase per winding, because the 3 phases are out of phase with each other by 120 degs., a 'Rotating' magnetic field is set up. This magnetic field cuts across the bars of the squirrel cage to induce an alternating magnetic field that interacts with the main field. At start-up this rotor field frequency will be the same as the supply frequency, but as the rotor speeds up, this frequency reduces to one slightly lower than the supply. The difference between the two is the slip, usually expressed as proportional slip. The rotor can never run at the same speed as the supply frequency, as then the rotating magnetic field will not cut the rotor bars.



3. Switch ON the supply. Increase the variac output voltage gradually until rated

voltage is observed in voltmeter. Note that the induction motor takes large current

initially, so, keep an eye on the ammeter such that the starting current current

should not exceed 7 Amp.

4. By the time speed gains rated value, note down the readings of voltmeter,

ammeter, and wattmeter at no-load.

5. Now the increase the mechanical load by tightening the belt around the brake

drum gradually in steps.

6. Note down the various meters readings at different values of load till the ammeter

shows the rated current.

7. Reduce the load on the motor finally, and switch OFF the supply.

MODEL CALCULATIONS:

Input power drawn by the motor W = (W1 + W2) watts

Shaft Torque, Tsh = 9.81 (S1 ~ S2) R N-m R Radius of drum in mts.

Output power in watts = wattsTN sh

60

2

100% xwattsinpowerInput

wattsinpoweroutputefficiency

p

fxNwherex

N

NNslip s

s

s 120100%

power factor of the induction motor LL IV

W

3cos



MODEL GRAPHS:

1. Speed or slip Vs output power

2. Torque Vs output power

3. % efficiency Vs output power

OBSERVATIONS:

S.

No.

V

(Volts)

I

(Amps)

Power,

W

(Watts)

Speed

(RPM)

Torque

(N-m)

Spring

balance

(Kg)

% Slip Cos Ø Output

Power

(W)

%η

W

1

W2 S1 S2



PRECAUTIONS:



VIVA Questions:

1. Why starter is used? What are different types of starters?

2. Compare a slip ring induction motor with cage induction motor?

3. Why the starting torque is zero for a Single Phase induction motor and non-zero

of 3phase induction motor?

4. What are the disadvantages of this method?

5. Can we use rotor resistance method for starting?

RESULT:

CONCLUSION:



SEPARATION OF NO LOAD LOSSES IN 1-Φ

TRANSFORMER

AIM:

To separation the Eddy current loss and Hysteresis loss from the iron loss of 1-Φ

transformer.

APPARATUS REQUIRED:

Sl.



2 Ammeter MC (0-2.5)A 1 no

3 Rheostat Wire-wound

350 Ω /1.7A

350Ω /1.7A

1 no

2 no


5 Wattmeter Electro dynamo meter type 10A/600V LPF 1 no


NAME PLATE DETAILS:

Rating

Primary Voltage

Secondary Voltage



CIRCUIT DIAGRAM:

PROCEDURE:

1. Make the circuit connections as per the circuit diagram.

2. The prime mover is started with the help of 3-point starter and it is made to run at

rated speed.

THEORY:

When a transformer is operated at no load, power drawn from the supply is equal to the no load losses, which are equal to the sum of constant losses (iron losses) and copper losses in the primary winding. The no load current of the transformer is quite small (of the order of 5 percent of full load current) and the resistance of the primary winding is also low, as such copper losses in the primary winding under this condition is negligible. Hence, the power drawn from the supply under no load condition can be approximately taken as the total iron losses.

Iron losses = Hysteresis loss + Eddy current loss

Hysteresis loss propoertional to Bn f which is proportional to f, with flux density being kept constant

= K1f



3. By varying alternators field rheostat gradually, the rated primary voltage is

applied to transformer.

4. By adjusting the speed of prime mover the required frequency, is obtained and

corresponding reading are noted.

5. The experiment is repeated for different frequency and corresponding readings are

tabulated.

6. The prime mover is switched off using the DPIC switch after bringing all the

rheostats to initial position

7. From the tabulated readings the iron loss is separated from eddy current loss and

hysteresis loss by using respective formulae.

OBSERVATIONS:

Separation of No load losses in single phase Transformer:

Multiplication factor=

S.No Speed of

the prime

mover

N(rpm)

Supply

frequency

(f)Hz

Primary

voltage

(V)volts

Wattmeter

readings(w)

Iron or core

loss

(Wi)watts

Wi/f

Observed

(watts)

Actual

(watts)

CALCULATIONS:

1. Frequency(f)=PNs/120

Where P-number of poles; Ns-Synchronous speed in rpm

2. Hysteresis loss(Wh)=Af

3. Eddy current loss(We)=Bf2

4. Iron loss or core loss(Wi)= We +Wh



MODEL GRAPH:

The graph drawn as frequency Vs( Wi/f)

PRECAUTIONS:

1. The motor field rheostat should be kept at minimum resistance position.

2. The alternator field rheostat should be kept maximum resistance position.

3. The motor should be run in anticlockwise direction.

4. Avoid loose connections.

5. Take the readings with any parallax error.

Before making or breaking the circuit, supply must be switched off6.



RESULT:

CONCLUSION:

VIVA QUESTIONS:

1. What are core losses in a transformer? Why they occur? On what factors do they

depend? What are the usual methods that are being employed in reducing them?

2. How does change in frequency affect the operation of a given transformer?

3. A transformer is designed for 50C/S operation. It is worked at double and half the

designed frequency what changes do you except in the performance? Discuss?

4. Whether you can excite a transformer from a DC supply of rated voltage Justify

your answer



DETERMINATION OF Xd AND Xq OF SALIENT POLE

SYNCHRONOUS MOTOR

AIM:

To determine the direct axis reactance Xd and quadrature axis reactance Xq by

conducting a slip test on a salient pole synchronous machine.

APPARATUS REQUIRED:

Sl.




3 Rheostat Wire-wound 350 Ω /1.7A 1 no



NAME PLATE DETAILS

3- ɸ Synchronous motor

Power Rating:

Rated voltage

rated current:

Speed



CIRCUIT DIAGRAM:

PROCEDURE:


2. Initially set field regulator, 3-ɸ variac at minimum position and TPST switch

open.

3. The DC motor is started slowly by sliding starter handle and it is run at a speed

slightly less than the synchronous speed of the alternator.

4. Close the TPST switch.

5. With field winding left open, a positive sequence balanced voltages of reduced

magnitude (around 25% of rated Value) and of rated frequency are

impressed across the armature terminals.

THEORY:

A Salient Pole Synchronous Generator, is distinguished from a round rotor machine by constructional features of field poles which project with a large interpolar air gap. This type of construction is commonly employed in machines coupled to hydroelectric turbines which are inherently slow-speed ones so that the Salient Pole Synchronous Generator has multiple pole pairs as different from machines coupled to high-speed steam turbines (3,000/1,500 rpm) which have a two- or four-pole structure. Salient pole machine analysis is made through the two-reaction theory outlined below.



6. The prime mover (DC motor) speed is adjusted till ammeter and voltmeters

pointers swing slowly between maximum and minimum positions.

7. Under this condition , readings of maximum and minimum values of both

ammeter and voltmeter are recorded

CALCULATIONS:

Maximum armature terminal voltage per phase Xd=

Minimum armature current per phase

Xq =

Note:

1. When performing this test, the slip should be made as small as possible.

2. During Slip test, it is observed that swing of the ammeter pointer is very wide,

whereas the voltmeter has only small swing.

TABULAR COLUMN:

Sl no. Speed Vmax

(VL)

Vmin

(VL)

Imax

(IL)

Imin

(IL)

Xd Xq

RESULT:

CONCLUSION:

Minimum armature terminal voltage per phaseMaximum armature current per phase



VIVA QUESTIONS:

1. Why ASA method is superior to ZPF method?

2. Why do you conduct Slip-test?

3. Describe voltage regulation of an alternator?

4. What is another name of Potier Triangle method?

5. What are the different methods used to find out the regulation. Compare them?

6. What is the reaction theory and what is its significance?

7. How do you calculate Synchronous impedance using OCC and SC tests on a synchronous machine?

8. What is the use of Damper Windings?

9. What are the different methods of starting a synchronous motor?

10. The armature winding of an alternator is in star or delta or both?



VIVA QUESTIONS AND ANSWERS

1. What is magnetic circuit?

The closed path followed by magnetic flux is called magnetic circuit

2. Define magnetic flux?

The magnetic lines of force produced by a magnet is called magnetic flux it is denoted as Ф

and its unit is Weber

3. Define magnetic flux density?

It is the flux per unit area at right angles to the flux it is denoted by B and unit is

Weber/metersquare

4. Define magneto motive force?

MMF is the cause for producing flux in a magnetic circuit. the amount of flux setup in the

core decent upon current(I)and number of turns(N).the product of NI is called MMF and it

determine the amount of flux setup in the magnetic circuit

MMF=NI ampere turns (AT)

5. Define reluctance?

The opposition that the magnetic circuit offers to flux is called reluctance. It is defind as the

ratio of MMF to flux. It is denoted by S and its unit is AT/m

6. What is retentivity?

The property of magnetic material by which it can retain the magnetism even after the

removal of inducing source is called retentivity

7. Define permeance?

It is the reciprocal of reluctance and is a measure of the cause the ease with which flux

can pass through the material its unit is wb/AT

8. Define magnetic flux intensity?

It is defined as the mmf per unit length of the magnetic flux path. it is denoted as H and its

unit is AT/m

H=NI/L



9. Define permeability?

Permeability of a material means its conductivity for magnetic flux. Greater the

permeability of material, the greaters its conductivity for magnetic flux and vice versa

10. Define relative permeability? It is equal to the ratio of flux density produced in that

material to the flux density

produced in air by the same magnetizing force

μr=μ/μ0

11. What is mean by leakage flux?

The flux does not follow desired path in a magnetic circuit is called leakage flux

12. What is leakage coefficient?

Leakage coefficient=total flux/useful flux

13. State faradays law of electromagnetic induction

Whenever a flux linking in the coil changes emf always induced in the conductor the

magnitude of induced emf is proportional to rate of change flux linkage

e = NdФ/dt

14. State Lenz law?

The law states that induced emf always opposite to applied voltage source

15. Define self inductance?

The property of a coil that opposes any change in the amount of current flowing through it is

called self inductance

16. Define mutual inductance?

The property of a coil to produce emf in a coil due to change in the value of current or flux in

it is called mutual inductance

17. Define coefficient coupling?

It is defined as the fraction of magnetic flux produced by the current in one coil that links

the other coil

18. Give the expression for hysteresis loss and eddy current loss?



19. What is dynamically induced emf?

An induced emf is produced by the movement of the conductor in a magnetic field. this emf is

called dynamically induced emf. The dynamically induced emf

e =BlvsinӨ

20. What is fringing effect?

It is seen that the useful flux passing across the air gap tends to buldge outwords, there by

increasing the effective area of the air gap and reducing the flux density in the gap is called

fringing effect

21. State two types of IM?

1. Squirrel cage IM

2. Slip ring IM

22. State ohms law for magnetic circuits?

Ohms law for magnetic circuits mmf=flux X reluctance

23. What is statically induced emf?

Conductor is stationary and the magnetic field is moving or changing the induced emf is

called stationary induced emf

24. How eddy current losses are minimized?

By laminating the core’

25. State types of electrical machines?

1.DC machines 2.AC machines 3.Special machines

26. What is mean by stacking factor?

Magnetic cores are made up of thin, lightly insulated laminations to reduce the eddy current

loss. As a result, the net cross sectional area of the core occupied by the magnetic material is

less than its gross cross section; their ratio being is called the stacking factor. The stacking

value is normally less than one .its value vary from 0.5 to 0.95 .the stacking factor value is

also reaches to one as the lamination thickness Increases

27. What are the magnetic losses?

1. Eddy current loss 2. Hysterisis loss



28. Types of induced emf?

1. Dynamically induced emf 2. Statically induced emf

1. Define a transformer?

A transformer is a static device which changes the alternating voltage from one level to

another. 2 What is the turns ratio and transformer ratio of transformer?

Turns ratio = N2/ N1 Transformer = E2/E1 = I1/ I2 =K

3. Mention the difference between core and shell type transformers?

In core type, the windings surround the core considerably and in shell type the core

surrounds the windings i.e winding is placed inside the core

4. What is the purpose of laminating the core in a transformer?

In order to minimise eddy current loss.

5. Give the emf equation of a transformer and define each term?

Emf induced in primary coil E1= 4.44fФmN1 volt Emf induced in secondary Coil E2 =4.44

fФmN2. f-----------freq of AC input Ф---------maximum value of flux in the core N1, N2----

Number of primary & secondary turns.

6. Does transformer draw any current when secondary is open? Why?

Yes, it (primary) will draw the current from the main supply in order to magnetize the core

and to supply for iron and copper losses on no load. There will not be any current in the

secondary since secondary is open.

7. Define voltage regulation of a transformer?

When a transformer is loaded with a constant primary voltage, the secondary voltage

decreases for lagging PF load, and increases for leading PF load because of its internal

resistance and leakage reactance. The change in secondary terminal voltage from no load to

full load expressed as a percentage of no load or full load voltage is termed as regulation.

%regulation =E2-V2/E2 *100 V2>E2 for leading p.f load V2<E2 for lagging p.f load

8. Define all day efficiency of a transformer?

It is computed on the basis of energy consumed during a certain period, usually a day of 24

hrs. All day efficiency=output in kWh/input in kWh for 24 hrs.



9. Why transformers are rated in kVA?

Copper loss of a transformer depends on current & iron loss on voltage. Hence total losses

depend on Volt-Ampere and not on PF. That is why the rating of transformers is in kVA and

not in kW.

10. What determines the thickness of the lamination or stampings?

1.Frequency 2.Iron loss

11. What are the typical uses of auto transformer?

1.To give small boost to a distribution cable to correct for the voltage drop. 2. as induction

motor starter.

12. What are the applications of step-up & step-down transformer?

Step-up transformers are used in generating stations. Normally the generated voltage will be

either 11kV. This voltage (11kV) is stepped up to 110kV or 220kV or 400Kv and transmitted

through transmission lines (simply called as sending end voltage). Step-down transformers

are used in receiving stations. The voltage are stepped down to 11kV or 22kV are stepped

down to 3phase 400V by means of a distribution transformer and made available at consumer

premises. The transformers used at generating stations are called power transformers.

13. How transformers are classified according to their construction?

1. Core type 2.shell type. In core type, the winding (primary and secondary) surround

the core and in shell type, the core surround the winding.

14. Explain on the material used for core construction?

The core is constructed by sheet steel laminations assembled to provide a continuous

magnetic path with minimum of air gap included. The steel used is of high silicon content

sometimes heat treated to produce a high permeability and a low hysteresis loss at the usual

operating flux densities. The eddy current loss is minimized by laminating the core, the

laminations being used from each other by light coat of coreplate vanish or by oxide layer on

the surface. The thickness of lamination varies from 0.35mm for a frequency of 50Hz and

0.5mm for a frequency of 25Hz.

15. How does change in frequency affect the operation of a given transformer?

With a change in frequency, iron and copper loss, regulation, efficiency & heating varies

so the operation of transformer is highly affected.



16. What is the angle by which no-load current will lag the ideal applied voltage?

In an ideal transformer, there are no copper & core loss i.e. loss free core. The no load current

is only magnetizing current therefore the no load current lags behind by angle 900. However

the winding possess resistance and leakage reactance and therefore the no load current lags

the applied voltage slightly less than 900 .

17. List the arrangement of stepped core arrangement in a transformer?

1. To reduce the space effectively 2. To obtain reduced length of mean turn of the winding 3.

To reduce I2R loss.

18. Why are breathers used in transformers?

Breathers are used to entrap the atmospheric moisture and thereby not allowing it to pass on

to the transformer oil. Also to permit the oil inside the tank to expand and contract as its

temperature increases and decreases.

19. What is the function of transformer oil in a transformer?

1. It provides good insulation 2. Cooling.

20. Can the voltage regulation goes –ive? If so under what condition?

Yes, if the load has leading PF.

21. Distinguish power transformers & distribution transformers?

Power transformers have very high rating in the order of MVA. They are used in generating

and receiving stations. Sophisticated controls are required. Voltage ranges will be very high.

Distribution transformers are used in receiving side. Voltage levels will be medium. Power

ranging will be small in order of kVA. Complicated controls are not needed.

22. Name the factors on which hysteresis loss depends?

1. Frequency 2. Volume of the core 3. Maximum flux density

23. Why the open circuit test on a transformer is conducted at rated voltage?

The open circuit on a transformer is conducted at a rated voltage because core loss depends

upon the voltage. This open circuit test gives only core loss or iron loss of the transformer.



24. What is the purpose of providing Taps in transformer and where these are provided?

In order to attain the required voltage, taps are provided, normally at high voltages

side(low current).

25. What are the necessary tests to determine the equivalent circuit of the transformer?

1. Open circuit test 2. Short circuit test

26. Define efficiency of the transformer?

Transformer efficiency ƞ= (output power/input power) x 100

27. Mention the difference between core and shell type transformers?

In core type, the windings surrounded the core considerably and in shell type the core

surround the windings i.e winding is placed inside the core

28. Full load copper loss in a transformer is 1600W. What will be the loss at half load?

If n is the ratio of actual load to full load then copper loss = n2 (F.L copper loss) Pc = (0.5)2

– 1600=400W.

29. Define all day efficiency of a transformer?

It is computed on the basis of energy consumed during a certain period, usually a day of

24 hrs. All day efficiency=output in kWh/input in kWh tor 24 hrs.

30. List the advantage of stepped core arrangement in a transformer?

1.To reduce the space effectively 2.To obtain reduce length of mean turn of the winding 3.To

reduce I2R loss.

31. Why are breathers used in transformers?

Breathers are used to entrap the atmospheric moisture and thereby not allowing it to pass on

to the transformer oil. Also to permit the oil inside the tank to expand and contract as its

temperature increases and decreases

1. State the principle of electromechanical energy conversion?

The mechanical energy is converted in to electrical energy which takes place through either

by magnetic field or electric field



2. Distinguish between statically induced emf and dynamically induced emf?

When emf induced in a conductor is stationary in a magnetic field then we call it statically

induced emf. If emf is induced in a conductor due to relative motion between conductor and

the field then it call it as dynamically induced emf.

3. What does speed voltage mean?

It is that voltage generated in that coil, when there exists a relative motion between coil and

magnetic field

4. Give example for single and multiple excited systems?

Single excited system-reluctance motor, single phase transformer, relay coil Multiply

excited system-alternator, electro mechanical transducer

5. Why do all practical energy conversion devices make use of the magnetic field as a

coupling medium rather than electric field?

When compared to electric field energy can be easily stored and retrieved form a

magnetic system with reduced losses comparatively. Hence most all practical energy

conversion devices make use of magnetic medium as coupling 6. State necessary condition for

production of steady torque by the interaction of stator and rotor field in electric machines?

1.The stator and rotor fields should not have any relative velocity or speed between each other

2.Airgap between stator and rotor should be minimum 3.Reluctance of iron path should be

negligible 4.Mutual flux linkages should exist between stator and rotor windings

7. Write the application of single and doubly fed magnetic systems?

Singly excited systems are employed for motion through a limited distance or rotation through

a prescribed angle Whereas multiply excited systems are used where continues energy

conversion take place and in ease of transducer where one coil when energized the care of

setting up of flux and the other coil when energized produces a proportional signal either

electrical or mechanical

8. Explain the following with respect to rotating electrical machines

1. Pole pitch 2. Charding angle 1. Pole pitch is that centre to centre distance between any two

consecutive poles in a rotating machine, measured in slots per poles 2. Chording angle is that

angle by which the coil span is short of full pitched in electrical degrees



9. Why energy stored in a magnetic material always occur in air gap

In iron core or steel core the saturation and aging effects form hindrance to storage Built in air

gap as reluctance as well permeability is constant, the energy storage takes place linearly

without any complexity. Hence energy is stored in air gap in a magnetic medium

10. What is the significance of co energy?

When electrical energy is fed to coil not the whole energy is stored as magnetic energy .the co

energy gives a measure of other energy conversion which takes place in coil then magnetic

energy storage

1. Field energy 2. Coenergy

11. Write the equation which relates rotor speed in electrical and mechanical radians per

second?

ὠe=ὠm(p/2) ὠe=speed in electrical radians per sec ὠm=speed in mechanical radians per sec

p=no of poles

12. Relate co energy density and magnetic flux density?

Co energy density=wf=∫0λ (I, x) di

wf=1/2BH

13. Short advantages of short pitched coil?

1. Hormonics are reduced in induced voltage 2. Saving of copper 3. End connections are shorter

14. What is the significance of winding factor?

Winding factor gives the net reduction in emf induced due to short pitched coil wound in

distributed type

Winding factor kw=kpkd kp= pitch factor kd= distribution factor kp= cos(α/2) kd=

sin(mγ/2)/msin(γ/2)

15. What is the necessity to determine the energy density in the design of rotating

machines?

Energy density wf=B2/2μ



16. Derive the relation between co energy and the phase angle between the rotor and

stator fluxes of the rotating machines? F1, f2 are the rotor and stator flux peak values

respectively

17. Write the energy balance equation for motor?

Mechanical energy o/p-=electrical energy i/p-increase in field energy Ff dx=idλ-dWf



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